Preparing NOJ

A+B Problem

1000ms 65536K

Description:

Given two integers $$$a$$$ and $$$b$$$, calculate their sum.

Input:

Two integers $$$a$$$ and $$$b$$$, satisfying $$$-32767 \le a, b \le 32767$$$.

Output:

An integer, denotes the sum of $$$a$$$ and $$$b$$$.

Sample Input:

1 2

Sample Output:

3

Note:

Q: Where are the input and the output?

A: Your program shall always read input from stdin (Standard Input) and write output to stdout (Standard Output). For example, you can use scanf in C or cin in C++ to read from stdin, and use printf in C or cout in C++ to write to stdout.

You shall not output any extra data to standard output other than that required by the problem, otherwise you will get a Wrong Answer.

User programs are not allowed to open and read from/write to files. You will get a Runtime Error or a Wrong Answer if you try to do so.

Here is a sample solution using C++/G++ :

#include <iostream>

using namespace std;

int main()
{
    int a,b;
    cin >> a >> b;
    cout << a+b << endl;
    return 0;
}

It's important that the return type of main() must be int when you use G++/GCC, or you may get compile error.

Here is a sample solution using C/GCC :

#include <stdio.h>

int main()
{
    int a,b;
    scanf("%d %d",&a, &b);
    printf("%d\n",a+b);
    return 0;
}

Here is a sample solution using Java :

import java.io.*;
import java.util.*;
public class Main
{
    public static void main(String args[]) throws Exception
    {
        Scanner cin=new Scanner(System.in);
        int a=cin.nextInt(),b=cin.nextInt();
        System.out.println(a+b);
    }
}

With one of the most elegant languages in the world, PHP could do that within one line :

<?=array_sum(fscanf(STDIN, "%d %d"));

Here is a sample solution using Python2

print sum(int(x) for x in raw_input().split(' '))

Here is a sample solution using Python3

print(sum(int(x) for x in input().split(' ')))

Here is a sample solution using Go

package main
import "fmt"
func main() {
    var a, b int
    fmt.Scanf("%d%d", &a, &b)
    fmt.Printf("%d\n", a + b)
}

Here is a sample solution using C#

using System;
using System.Linq;

class Program {
    public static void Main(string[] args) {
        Console.WriteLine(Console.ReadLine().Split().Select(int.Parse).Sum());
    }
}

Here is a sample solution using Javascript

const [a, b] = readline().split(' ').map(n => parseInt(n, 10));
print((a + b).toString());

Here is a sample solution using Ruby

a, b = gets.split.map(&:to_i)
puts(a + b)

Here is a sample solution using Rust

use std::io;
 
fn main() {
    let mut line = String::new();
    io::stdin().read_line(&mut line).expect("stdin");
 
    let sum: i32 = line.split_whitespace()
                       .map(|x| x.parse::<i32>().expect("integer"))
                       .sum(); 
    println!("{}", sum);
}

Here is a sample solution using Haskell

main = print . sum . map read . words =<< getLine

Here is a sample solution using Free Pascal

var a, b:longint;
begin
    readln(a, b);
    writeln(a + b);
end.

Here is a sample solution using Free Basic

Dim a As Integer
Dim b As Integer

Open Cons For Input As #1
Input #1, a, b
Print Str(a+b)

Here is a sample solution using Wenyan, of which you may ask, why I am seeing this?

施「require('fs').readFileSync」於「「/dev/stdin」」。名之曰「數據」。
施「(buf => buf.toString().trim())」於「數據」。昔之「數據」者。今其是矣。
施「(s => s.split(' '))」於「數據」。昔之「數據」者。今其是矣。
注曰。「「文言尚菜,無對象之操作,故需 JavaScript 之语法」」。

夫「數據」之一。取一以施「parseInt」。名之曰「甲」。
夫「數據」之二。取一以施「parseInt」。名之曰「乙」。

加「甲」以「乙」。書之。

Here is a sample solution using Text :

WAAAAAAAIT A MINUTE, THERE IS NONE! STOP DAYDREAMING.

Info

NOJ

Provider NOJ

Code NOJ1001

Tags

Submitted 2751

Passed 1311

AC Rate 47.66%

Date 08/20/2021 19:49:10

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